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Homework Help: Choosing the Resistance in a Photoconductive Circuit

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data
    What should I choose for the resistor [itex]R_{F}[/itex] and capacitor [itex]C_{F}[/itex] in this circuit?


    I have an unmounted silicon photodiode, with specs:

    Reverse Voltage = 100 [itex]V[/itex]
    Short Circuit Current = 50 [itex]\mu A[/itex] (typical)
    Dark Current = 0.5 [itex]n A[/itex] (typical), 1.0 [itex]n A[/itex] (max)
    Shunt Resistance = 250 [itex]M \Omega[/itex] (min), 500 [itex]M \Omega[/itex] (typical)
    Junction Capacitance = 8 [itex]pF[/itex] (typical)
    Breakdown Voltage = 75 [itex]V[/itex] (typical)

    I'm trying to figure out how to properly use it. Laser light from an ~800nm diode laser will be incident on it, and I would like to view its output signal on an oscilloscope via a BNC coaxial cable.

    I've learned that I can operate the photodiode in "photovoltaic mode", where there is zero bias and low dark current, but slow response time, or "photoconductive mode", where there is a negative bias and fast response time, but higher dark current.

    Due to the circumstances of my experiment, I'm choosing to go with "photoconductive mode". Hence, the above circuit has a negative bias. I got it from AP Technologies website, where I also saw some useful formulas:

    2. Relevant equations
    [tex]V_{OUT} = I_{S} R_{F}[/tex]

    where [itex]I_{S}[/itex] is the photodiode light signal current and [itex]R_{F}[/itex] is the feedback resistance.

    [tex]C_{F} = \frac{1}{2} \pi f R_{F}[/tex]

    where [itex]C_{F}[/itex] is the feedback capacitance and [itex]f[/itex] is the maximum operating frequency.

    3. The attempt at a solution
    I guess I'm just not sure what to use for values for calculating the resistance and capacitance needed. I've searched around online, and even though I've found some useful documents on this, I'm still left with two questions:

    1. How large of a resistor do I need to not fry the diode?

    2. What capacitor should I use? What is the "maximum operational frequency" referring to? Is it related to the RC time constant, or to the frequency of the incident light, etc?

    My naive guess for the first one would be to use [itex]V_{OUT} \sim 10 V[/itex] and [itex]I_{S} \sim 10 \mu A[/itex] so that

    [tex]R_{F} = \frac{10 V}{10 \mu A} = 1 M \Omega[/tex]

    That's big, though!
  2. jcsd
  3. Jun 16, 2009 #2


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    RF will depend on what current you expect for IS, and how much signal you would like to have. IS will depend on the amount of power from the diode beam hitting the photodiode's active area. You might take 1V as a reasonable signal level.

    This circuit might work fine with no capacitor CF. But if you use one, the question to answer is how fast must the output signal respond? Choose CF so that the RC time constant is 1/3 of that time or less.
  4. Jun 17, 2009 #3
    Thanks Redbelly98. Ignoring the capacitor, then, I'm still slightly confused as to what 'spec' value I should be looking at to help me choose a resistor that would not fry the diode.

    Maybe my concern is unwarranted. In reverse-biasing the diode, we're not causing additional current flow through the diode, we're just setting up a potential difference, and the result is more output current from the photodiode? And it's current that would, in my words, "fry" the photodiode.

    Since the capacitor-free circuit would just be an inverting amplifier, for which:

    [tex]V_{out} = - \frac{R_{F}}{R_{D}} V_{in}[/tex]

    where [itex]R_{D}[/itex] is the resistance of the diode, then with Ohm's law we could write:

    [tex]I_{S} R_{D} = -\frac{R_{D}}{R_{F}} V_{out} \rightarrow R_{F} = -\frac{V_{out}}{I_{S}}[/tex]

    and with [itex]I_{S} = 50 \mu A[/itex] and [itex]V_{out} = 1 V[/itex] we have

    [tex]R_{F} = 50 M \Omega[/tex]

    Is that roughly the right way to think about it?
  5. Jun 17, 2009 #4


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    1V / 50 μA is 20kΩ, not 50MΩ. 20kΩ sounds reasonable.

    Forget about "resistance of the diode", instead think in terms of diode current. Actually, this is essentially what you are doing since your RD terms cancelled out of the equation.

    The photodiode is essentially a current source here. At 800 nm, it can generate 0.5-1.0 Amps per Watt of optical power. To generate 50 μA, it would be receiving 50-100 μW from the laser. This should not hurt the photodiode.

    I'll also suggest building a preliminary circuit and experimenting to find the right resistor value. No op-amp, just a loop circuit containing the photodiode, biasing voltage, and resistor. A simple 1.5V battery (or two of them) can be used for bias. Then try different resistor values until you get a volt or so of signal across the resistor.

    p.s. if the resistor voltage is close to or larger than the bias voltage, then you are saturating the photodiode. Increase the bias voltage or reduce the resistor so that the resistor voltage is less than the bias.

    p.p.s. For really fast-changing signals, say in the RF frequency range (several ns or faster), it is common practice to use a 50Ω resistor with the photodiode, and then use amplifiers to amplify the voltage across the 50Ω resistor.
  6. Jun 17, 2009 #5
    Oops. Right, that makes more sense.

    Thanks! This really clears things up. I'll try your suggestion of circuit-sans-op-amp, and go from there.
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